Question

Помогите решить мервые 2 номера пожалуйста, даю 100 баллов

Answer 1

1. a) $sin^{2}\alpha$ + $cos^{2} \alpha$ = 1

$sin^{2}\alpha$ = 1 - $cos^{2} \alpha$ = 1 - $(-1/4)^{2}$ = 1 - 1/16 = 15/16

|sin $\alpha$| = $\sqrt{\frac{15}{16} }$ = 1/4 * $\sqrt{15}$

б) $sin^{2}\alpha$ + $cos^{2} \alpha$ = 1

$cos^{2} \alpha$ $\alpha$ = 1 - $sin^{2} \alpha$ = 1 - $(-2/3)^{2}$ = 1 - 4/9 = 5/9

|cos $\alpha$| = $\sqrt{\frac{5}{9} }$ = 1/3 * $\sqrt{5}$

в) $sin^{2}\alpha$ + $cos^{2} \alpha$ = 1

$sin^{2}\alpha$ = 1 - $cos^{2} \alpha$ = 1 - $1-(\sqrt{3} /2)^{2}$ = 1 - 3/4 = 1/4

sin $\alpha$ = $\sqrt{\frac{1}{4} }$ = $\frac{1}{2}$

|tg $\alpha$| = sin $\alpha$ / cos $\alpha$ = (1/2) / ($\sqrt{3}$/2) = 1 / $\sqrt{3}$

2. a) $(\frac{1}{3})^{2}$ + $(2\sqrt{3} / 3)^{2}$ = 1/9 + 8/9 = 1     Лежит.

б) $(\sqrt{3} / 2)^{2}$ + $(\sqrt{3} / 2)^{2}$ = 3/4 + 3/4 = 6/4 $\neq$ 1     Не лежит.

в) $2^{2}$ + $3^{2}$ = 4 + 9 = 13 $\neq$ 1     Не лежит.